Author Topic: Intergration problem (Read 1609 times) Tweet Share

TrueTears · « **on:** November 27, 2008, 06:22:19 pm »

1. Suppose that a point moves along some unknown curve y= f(x) in such a way that at each point (x,y) on the curve the tangent line has slope x^2. Find an equation for the curve, given that it passes through (2,1).

2. Find the equation of the curve which passes through the point (-1,2) and has the property that for each point (x,y) on the curve the gradient equals the square of the distance between the point and the y-axis.

3. Given $\int_{2}^{5} f(x)\, dx$ = 12 evaluate $\int_{2}^{4} f(x)+4\, dx$ + $\int_{4}^{5} f(x)\, dx$

bucket · « **Reply #1 on:** November 27, 2008, 06:46:32 pm »

Hmm, I'll have a shot at number two... It's been a few weeks though..
The condition is that $m=x^2$ for every point.
We know that the gradient of a curve is it's derivative, therefore we know that the derivative of this curve in particular, at the point $(-1,2)$ will be $x^2$ , so we find the antiderivative of $x^2$ and plug in those coordinates to find $c$ :

$\int x^2 dx = \frac{1}{3}x^3+c$

$(-1,2)$
$2=\frac{1}{3}\cdot (-1)^3+c$
$2=-\frac{1}{3}+c$
$\frac{7}{3}=c$

Therefore, the equation is $\frac{1}{3}x^3+\frac{7}{3}$

I hope that is right, lol.

methodsboy · « **Reply #2 on:** November 27, 2008, 07:10:21 pm »

uuuggh
im so over methods......

dekoyl · « **Reply #3 on:** November 27, 2008, 08:14:41 pm »

3.
$\int_{2}^{4} f(x) + 4.dx + \int_{4}^{5}f(x).dx$

$\int_{2}^{5} f(x).dx + \int_{2}^{4}4.dx$

$12 + [4x]_{2}^{4}$

$= 12 + 8 = 20$

I think..

Mao · « **Reply #4 on:** November 28, 2008, 12:59:18 am »

1. the way i interpreted the question is given $f'(x)=x^2$ and f(x) passes through (2,1), find f(x)

Flaming_Arrow · « **Reply #5 on:** November 28, 2008, 01:15:09 am »

1.

$\frac {dy}{dx} = x^2$

$\int x^2 dx = \frac {x^3}{3} + c$

sub and x and y to find y intercept(c)

$1 = \frac {8}{3} + c$

$c = \frac {-5}{3}$

$y =\frac {x^3}{3} - \frac {5}{3}$

$\implies \boxed { y = \frac {1}{3} (x^3-5)}$

Mao · « **Reply #6 on:** November 28, 2008, 01:25:22 am »

Another method:

$f(x)=\int_2^x t^2\; dt + 1$

TrueTears · « **Reply #7 on:** November 28, 2008, 11:20:55 am »

ah cool thanks all

dekoyl · « **Reply #8 on:** November 28, 2008, 12:00:32 pm »

Quote from: TrueTears on November 28, 2008, 11:20:55 am

ah cool thanks all

Were the answers all correct? I was slightly stuck on some

So I'm interested.

TrueTears · « **Reply #9 on:** November 28, 2008, 07:26:11 pm »

Quote from: dekoyl on November 28, 2008, 12:00:32 pm

Quote from: TrueTears on November 28, 2008, 11:20:55 am
ah cool thanks all
Were the answers all correct? I was slightly stuck on some So I'm interested.

hehe yeap they were all good thanks yet again

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